A group of 10 students participate in the chess club, karate club, or neither.

Let event [tex]$A$[/tex] be: The student is in the chess club.
Let event [tex]$B$[/tex] be: The student is in the karate club.

One of these students is randomly selected. What is [tex]$P(A \mid B)$[/tex]?

A. [tex]\frac{2}{10} \approx 0.20[/tex]
B. [tex]\frac{4}{6} \approx 0.67[/tex]
C. [tex]\frac{2}{6} \approx 0.33[/tex]
D. [tex]\frac{8}{10} \approx 0.80[/tex]



Answer :

To determine the conditional probability [tex]\( P(A \mid B) \)[/tex], which is the probability that a student is in the chess club given that they are in the karate club, we apply the formula for conditional probability:

[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \][/tex]

Where:
- [tex]\(P(A \cap B)\)[/tex] is the probability that a student is in both the chess club and the karate club.
- [tex]\(P(B)\)[/tex] is the probability that a student is in the karate club.

According to the problem statement, there are 10 students in total.

We know the following details:

1. There are 2 students in both the chess club and the karate club.
2. There are 4 students who are only in the chess club.
3. There are 2 students who are only in the karate club.
4. The remaining 2 students are in neither club.

This means:

- The number of students in both clubs ([tex]\(A \cap B\)[/tex]) is 2.
- The total number of students in karate club ([tex]\(B\)[/tex]) is the sum of those in both clubs and those only in karate:
[tex]\[ \text{Students in karate club} = 2 \ (\text{both clubs}) + 2 \ (\text{only karate}) = 4 \ (\text{total in karate}) \][/tex]

Substituting these values into the conditional probability formula:

[tex]\[ P(A \mid B) = \frac{\text{number of students in both clubs}}{\text{number of students in karate club}} = \frac{2}{6} \approx 0.33 \][/tex]

Thus, the correct answer is:

C. [tex]\(\frac{2}{6} \approx 0.33\)[/tex]